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I'm following Griffiths' intro QM text, 2nd edition. We have defined the angular momentum operator and obtained the commutation relations $[L_i, L_j] = i\hbar\epsilon_{ijk} L_k$. We notice in particular how the components are incompatible observables so there's no sense in trying to simultaneously diagonalising them. Instead, after noticing that $[L_i, L^2] = 0$, we try to simultaneously diagonalise $L^2$ and one component, say $L_z$.
We define the ladder operators $L_\pm = L_x + iL_y$, and observe that, since $[L^2, L_\pm] = 0$ and $[L_z, L_\pm] = \pm\hbar L_\pm$, if $\psi$ is a simultaneous eigenstate of $L^2, L_z$ with corresponding eigenvalues $\lambda,\mu$, then $L_\pm\psi$ is also a simultaneous eigenstate of $L^2, L_z$ with corresponding eigenvalues $\lambda, \mu\pm\hbar$.
Thus if one simultaneous eigenstate $\psi$ is known, from it we can obtain a whole sequence of simultaneous eigenstates by repeated application of the ladder operators. We observe however that for any given eigenstate $\lambda > \mu^2$, so that this method of construction of new states must fail at some point. We conclude that there must exist a "top" state $\psi_+$ (or "bottom" state $\psi_-$) such that $L_\pm\psi_\pm$, the result of raising/lowering it, cannot be normalisable.
At this point, Griffiths starts an argument to relate the eigenvalues $\lambda$ and $\mu$ (the conclusion is that $\mu_\pm = \pm\hbar l$ for some positive integer or half-integer $l$, and $\lambda=\hbar^2 l(l+1)$). The first argument presented relies on the assumption that $L_\pm \psi_\pm = 0$, which as we remarked earlier is not necessary—$L_\pm\psi_\pm$ need only be non-normalisable.
QUESTION: How to obtain the result without this unjustified assumption $L_\pm\psi_\pm = 0$? i.e. either justify it or assume only non-normalisability.
A footnote in the page mentions the fact that the assumption $L_\pm \psi_\pm = 0$ is not completely due, and refers the reader to Problem 4.18, which is said to explore this. Using
$$ L_\pm L_\mp = L^2 - L_z^2 \pm \hbar L_z \qquad\qquad L_\pm^\dagger = L_\mp $$
as suggested, I can obtain
$$ |L_\pm\psi|^2 = \<L_\pm\psi|L_\pm\psi\> = \lambda - \mu(\mu\pm\hbar) $$
but at this point Griffiths seems to assume $\lambda=\hbar^2 l(l+1)$ where $\mu_\pm=\pm\hbar l$ (which can then be used to conclude that $|L_\pm\psi_\pm|=0$ indeed), but that is cheating since these had been previously derived under the assumption we're trying to avoid.
